Question

In a series R-L circuit, the voltage of the battery is 10 V. Resistance and inductance are \( 10 \, \Omega \) and \( 10 \, \text{mH} \) respectively. Find the energy stored in the inductor when the current reaches \( \frac{1}{e} \) times of its maximum value.

• A. 0.67 mJ
• B. 1.33 mJ
• C. 0.33 mJ
• D. 0.50 mJ

Hint:

In R-L circuits, the energy stored in the inductor is related to the square of the current and depends on the time it takes to reach a fraction of its maximum value.

Answer 1

The Correct Option is A

Step 1: Formula for energy stored in the inductor.
The energy stored in an inductor is given by the formula \( E = \frac{1}{2} L I^2 \), where \( L \) is the inductance and \( I \) is the current. The current at a time \( t \) in an R-L circuit is given by \( I(t) = I_{\text{max}}(1 - e^{-t/\tau}) \), where \( \tau = \frac{L}{R} \). Step 2: Applying \( \frac{1}{e} \) of the maximum current.
Substitute \( I = \frac{I_{\text{max}}}{e} \) into the energy formula, considering the given values for \( L \) and \( R \). After performing the calculation, we get the energy stored as 0.67 mJ. Step 3: Conclusion.
The energy stored in the inductor is 0.67 mJ. Final Answer: \[ \boxed{0.67 \, \text{mJ}} \]